Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $x = \dfrac{a^2 + 8a + 15}{a^2 - 2a - 15} \times \dfrac{-3a + 15}{2a - 4} $
First factor out any common factors. $x = \dfrac{a^2 + 8a + 15}{a^2 - 2a - 15} \times \dfrac{-3(a - 5)}{2(a - 2)} $ Then factor the quadratic expressions. $x = \dfrac {(a + 3)(a + 5)} {(a + 3)(a - 5)} \times \dfrac {-3(a - 5)} {2(a - 2)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac { (a + 3)(a + 5) \times -3(a - 5)} { (a + 3)(a - 5) \times 2(a - 2)} $ $x = \dfrac {-3(a + 3)(a + 5)(a - 5)} {2(a + 3)(a - 5)(a - 2)} $ Notice that $(a + 3)$ and $(a - 5)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {-3\cancel{(a + 3)}(a + 5)(a - 5)} {2\cancel{(a + 3)}(a - 5)(a - 2)} $ We are dividing by $a + 3$ , so $a + 3 \neq 0$ Therefore, $a \neq -3$ $x = \dfrac {-3\cancel{(a + 3)}(a + 5)\cancel{(a - 5)}} {2\cancel{(a + 3)}\cancel{(a - 5)}(a - 2)} $ We are dividing by $a - 5$ , so $a - 5 \neq 0$ Therefore, $a \neq 5$ $x = \dfrac {-3(a + 5)} {2(a - 2)} $ $ x = \dfrac{-3(a + 5)}{2(a - 2)}; a \neq -3; a \neq 5 $